Chapter 9, Problem 9T

Find two unit vectors that are perpendicular to both j + 2k and i − 2j + 3k.

To determine

The two unit vectors that are perpendicular to both the vectors $j+2k$ and $i-2j+3k$.

Answer to Problem 9T

The two unit vectors perpendicular to both the vectors $j+2k$ and $i-2j+3k$ are $\langle \frac{7\sqrt{6}}{18},\frac{\sqrt{6}}{9},-\frac{\sqrt{6}}{18}\rangle$ and $\langle -\frac{7\sqrt{6}}{18},-\frac{\sqrt{6}}{9},\frac{\sqrt{6}}{18}\rangle$.

Explanation of Solution

Given:

The two vectors are given below,

$j+2k$

$i-2j+3k$

Formula used: The unit vector perpendicular to vector a and b is given below,

$u=\frac{a\times b}{\left|a\times b\right|}$ (1)

The vectors $u=\langle a_1,a_2,a_3\rangle$ and $v=\langle b_1,b_2,b_3\rangle$ are three-dimensional vectors, then the cross product of u and v is given below,

$u\times v=\left|\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$ (2)

The magnitude of a vector $v=a_{1}i+a_{2}j+a_{3}k$ is given below,

$\left|v\right|=\sqrt{a_1^2+a_2^2+a_3^2}$ (3)

Calculation:

Substitute 0 for $a_1$, 1 for $a_2$, 2 for $a_3$, 1 for $b_1$, $-2$ for $b_2$ and 3 for $b_3$ in equation (2), to find the vector perpendicular to both the vectors $j+2k$ and $i-2j+3k$,

$\begin{array}{c}a\times b=\left|\begin{array}{ccc}i& j& k\\ 0& 1& 2\\ 1& -2& 3\end{array}\right|\\ =\left|\begin{array}{cc}1& 2\\ -2& 3\end{array}\right|i-\left|\begin{array}{cc}0& 2\\ 1& 3\end{array}\right|j+\left|\begin{array}{cc}0& 1\\ 1& -2\end{array}\right|k\\ =\left(3-\left(-4\right)\right)i-\left(0-2\right)j+\left(0-1\right)k\\ =7i+2j-k\end{array}$

Substitute 7 for $a_1$, 2 for $a_2$ and $-1$ for $a_3$ in equation (3), to find the value of $\left|a\times b\right|$,

$\begin{array}{c}\left|a\times b\right|=\sqrt{{\left(7\right)}^2+{\left(2\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{49+4+1}\\ =\sqrt{54}\\ =3\sqrt{6}\end{array}$

Substitute $7i+2j-k$ for $a\times b$ and $3\sqrt{6}$ for $\left|a\times b\right|$ in equation (1), to find the unit vector perpendicular to both the vectors $j+2k$ and $i-2j+3k$,

$\begin{array}{c}u=\frac{7i+2j-k}{3\sqrt{6}}\\ =\frac{7}{3\sqrt{6}}i+\frac{2}{3\sqrt{6}}j-\frac{1}{3\sqrt{6}}k\\ =\frac{7\sqrt{6}}{18}i+\frac{2\sqrt{6}}{18}j-\frac{\sqrt{6}}{18}k\\ =\frac{7\sqrt{6}}{18}i+\frac{\sqrt{6}}{9}j-\frac{\sqrt{6}}{18}k\end{array}$

The unit vector $\frac{7\sqrt{6}}{18}i+\frac{\sqrt{6}}{9}j-\frac{\sqrt{6}}{18}k$ is perpendicular to both the vectors $j+2k$ and $i-2j+3k$.

Inverse the coefficients sign of the unit vector, to find the second unit vector perpendicular to both the vectors $j+2k$ and $i-2j+3k$.

$u=-\frac{7\sqrt{6}}{18}i-\frac{\sqrt{6}}{9}j+\frac{\sqrt{6}}{18}k$

Thus, the two unit vectors perpendicular to both the vectors $j+2k$ and $i-2j+3k$ are $\langle \frac{7\sqrt{6}}{18},\frac{\sqrt{6}}{9},-\frac{\sqrt{6}}{18}\rangle$ and $\langle -\frac{7\sqrt{6}}{18},-\frac{\sqrt{6}}{9},\frac{\sqrt{6}}{18}\rangle$.